Deal with overflow in exp using numpy

匿名 (未验证) 提交于 2019-12-03 09:05:37

问题:

Using numpy, I have this definition of a function:

def powellBadlyScaled(X):     f1 = 10**4 * X[0] * X[1] - 1     f2 = numpy.exp(-numpy.float(X[0])) + numpy.exp(-numpy.float(X[1])) - 1.0001     return f1 + f2 

This function is evaluated a huge number of times on an optimization routine. It often raises exception:

RuntimeWarning: overflow encountered in exp 

I understand that operand cannot be stored in allocated space for a float. But how can I overcome the problem?

回答1:

You can use the bigfloat package. It supports arbitrary precision floating point operations.

http://packages.python.org/bigfloat/

import bigfloat bigfloat.exp(5000,bigfloat.precision(100)) # -> BigFloat.exact('2.9676283840236670689662968052896e+2171', precision=100) 

Are you using a function optimization framework? They usually implement value boundaries (using penalty terms). Try that. Are the relevant values really that extreme? In optimization it's not uncommon to minimize log(f). (approximate log likelihood etc etc). Are you sure you want to optimize on that exp value and not log(exp(f)) == f. ?

Have a look at my answer to this question: logit and inverse logit functions for extreme values

Btw, if all you do is minimize powellBadlyScaled(x,y) then the minimum is at x -> + inf and y -> + inf, so no need for numerics.



回答2:

You can use numpy.seterr to control how numpy behaves in this circumstance: http://docs.scipy.org/doc/numpy/reference/generated/numpy.seterr.html

You can also use the warnings module to control how warnings are or are not presented: http://docs.python.org/library/warnings.html



回答3:

Maybe you can improve your algorithm by checking for which areas you get warnings (it will probably bellow certain values for X[ 0 ],X[ 1 ]), and replacing the result with a really big number. You need to see how your function behaves, I thing you should check e.g. exp(-x)+exp(-y)+x*y



回答4:

Depending on your specific needs, it may be useful to crop the input argument to exp(). If you actually want to get an inf out if it overflows or you want to get absurdly huge numbers, then other answers will be more appropriate.

def powellBadlyScaled(X):     f1 = 10**4 * X[0] * X[1] - 1     f2 = numpy.exp(-numpy.float(X[0])) + numpy.exp(-numpy.float(X[1])) - 1.0001     return f1 + f2   def powellBadlyScaled2(X):     f1 = 10**4 * X[0] * X[1] - 1     arg1 = -numpy.float(X[0])     arg2 = -numpy.float(X[1])     too_big = log(sys.float_info.max / 1000.0)  # The 1000.0 puts a margin in to avoid overflow later     too_small = log(sys.float_info.min * 1000.0)     arg1 = max([min([arg1, too_big]), too_small])     arg2 = max([min([arg2, too_big]), too_small])     # print('    too_small = {}, too_big = {}'.format(too_small, too_big))  # Uncomment if you're curious     f2 = numpy.exp(arg1) + numpy.exp(arg2) - 1.0001     return f1 + f2  print('\nTest against overflow: ------------') x = [-1e5, 0] print('powellBadlyScaled({}) = {}'.format(x, powellBadlyScaled(x))) print('powellBadlyScaled2({}) = {}'.format(x, powellBadlyScaled2(x)))  print('\nTest against underflow: ------------') x = [0, 1e20] print('powellBadlyScaled({}) = {}'.format(x, powellBadlyScaled(x))) print('powellBadlyScaled2({}) = {}'.format(x, powellBadlyScaled2(x))) 

Result:

Test against overflow: ------------ *** overflow encountered in exp  powellBadlyScaled([-100000.0, 0]) = inf powellBadlyScaled2([-100000.0, 0]) = 1.79769313486e+305  Test against underflow: ------------ *** underflow encountered in exp     powellBadlyScaled([0, 1e+20]) = -1.0001 powellBadlyScaled2([0, 1e+20]) = -1.0001 

Notice that powellBadlyScaled2 didn't over/underflow when the original powellBadlyScaled did, but the modified version gives 1.79769313486e+305 instead of inf in one of the tests. I imagine there are plenty of applications where 1.79769313486e+305 is practically inf and this would be fine, or even preferred because 1.79769313486e+305 is a real number and inf is not.



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